Question: Jovie is maintaining a camp fire. She has kept the fire steadily burning for $10$ hours with $15$ logs. She wants to know how many hours $(h)$ she could have kept the fire going with $9$ logs. She assumes all logs are the same. How many hours can Jovie keep the fire going with $9$ logs?
Answer: We can set up a proportion like this: $\dfrac{\text{Fire duration with 9 logs}}{\text{Fire duration with 15 logs}} = \dfrac{\text{9 logs}}{15\text{ logs}}$ Substituting values from the problem, we get this: $\dfrac{h\text{ hours}}{10\text{ hours}} = \dfrac{9\text{ logs}}{15\text{ logs}}$ Solve for $h$ : $\begin{aligned} \dfrac{h}{10} &= \dfrac{9}{15} \\\\\\ \dfrac{h}{10} &= \dfrac{3}{5} \\\\\\ h &= \dfrac{3}{5} \cdot 10 \\\\\\ h &= \dfrac{3\cdot 10}{5} \\\\\\ h &= \dfrac{3\cdot \stackrel{2}{\cancel{10}}}{\underset{1}{\cancel{5}}} \\\\\\ h &= \dfrac{6}{1}\\\\\\ h &= 6 \end{aligned}$ Jovie can keep the fire going for $6$ hours with $9$ logs. The answers: $\dfrac{h}{10} = \dfrac{9}{15}$ $6$ hours